Question

Question #0d9d2

Answer 1

The concentrations are [NH₄⁺] = 0.45 mol/L; [Cl⁻] = 0.45 mol/L; [NH₃] = 0.15 mol/L; [OH⁻] = 5.93 × 10⁻⁶ mol/L; [H₃O⁺] = 1.69 × 10⁻⁹ mol/L.

Step 1. Identify the species.

You have two equilibria here — the ionization of ammonia and the ionization of water.

The equations are:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
2H₂O = H₃O⁺ + OH⁻

The species involved in the equilibria are NH₃, NH₄⁺, OH⁻, and H₃O⁺.

You also have the Cl⁻ as a spectator ion from the NH₄Cl (5 species in total).

Step 2. Determine the concentrations.

You know immediately that

(1) [NH₃] = 0.15 mol/L

(2) [NH₄⁺] = 0.45 mol/L

(3) [Cl⁻] = o.45 mol/L

Since this is a buffer, we can use the Henderson-Hasselbalch Equation:

`"pOH" = "p"K_"b" + log((["NH"_4^+])/(["NH"_3])) = 4.75 + log((0.45 cancel("mol/L"))/(0.15 cancel("mol/L")))`

`= 4.75 + 0.48 = 5.23`

`["OH"^(-)] = 10^"-pOH" = 10^-5.23`

(4) `["OH"^(-)] = 5.93 × 10^-6 "mol/L"`

`["H"_3"O"^+] = K_w/(["OH"^(-)]) = (1.00 × 10^-14)/(5.93 × 10^-6)`

(5) `["H"_3"O"^+] = 1.69 × 10^-9 "mol/L"`

Answer 2

`[NH_(3(aq))]=0.15"mol/l"`

`[NH_(4(aq))^+]=0.45"mol/l"`

`[Cl_((aq))^-]=0.45"mol/l"`

`[H_((aq))^+]=1.69xx10^(-9)"mol/l"`

`[OH_((aq))^-]=5.93xx10^(-6)"mol/l"`

The mixture contains a lot of unreacted ammonia molecules and lots of ammonium ions.

Ammonium ions are slightly acidic:

`NH_(4(aq))^+rightleftharpoonsNH_(3(aq))+H_((aq))^+`

We can make some assumptions which means we don't have to go into working out equilibrium concentrations.

We can write an expression for `K_arArr`

`K_a=([NH_3][H^+])/([NH_4^+])=5.62xx10^(-10)"mol/l"`

The presence of ammonia in the mixture forces the equilibrium to the left.

This means we can assume the ammonium ion concentration is what we started with - `0.45"M"` and that the ammonia concentration is all due to the initial ammonia - `0.15"M"`

Putting in the numbers:

`5.62xx10^(-10)=(0.15xx[H^+])/(0.45)`

From which:

`[H^+]=1.69xx10^(-9)"mol/l"`

To find `[OH^-]` we use the ionic product of water:

`[H^+][OH^-]=10^(-14)mol^(2).l^(-2)"@"298K`

`[OH^(-)]=(10^(-14))/(1.69xx10^(-9))=5.93xx10^(-6)"mol/l"`

The chloride ions take no part in any of this so their concentrations remain at `0.45"mol/l"`.